# The Answers of Student Dawkins to a High School Math Quiz

Richard Dawkins has extensively discussed arithmetic. The theme of *The God Delusion* is that there is an arithmetical solution to the improbability of evolution in a one-off event, namely gradualism, whereas there is no arithmetical solution to the improbability of God. Obviously, the ‘improbability’ of God cannot be solved by gradualism.

It is encouraging that Richard Dawkins is interested in mathematics. If he were to learn to correct his mistakes in math, he might do very well in re-educating those whom he has deceived in mathematics, science and philosophy due to his errors in arithmetic.

The following is a math quiz based on problems in arithmetic addressed by Richard Dawkins and his answers, whether implicit or explicit, in his public work. I present this as a helpful perspective in the delineation of Dawkins’ public errors in arithmetic.

1) What is the opposite of +1?

Correct Answer: -1

Student Dawkins: Zero. Let us then take the idea of a spectrum of probabilities seriously between extremes of opposite certainty. The spectrum is continuous, but can be represented by the following seven milestones along the way:

Strong positive, 100%; Short of 100%; Higher than 50%; 50%; Lower than 50%; Short of zero; Strong negative, 0% (p 50-51. Ref 1).

Those, who aver that we cannot say anything about the truth of the statement, should refuse to place themselves anywhere in the spectrum of probabilities, i.e. of certitude. (p 51, Ref 1)

Critique of Dawkins’ answer:

On page 51 of The God Delusion, Dawkins devotes a paragraph to discussing the fact that his spectrum of certitude, from positive to its negative opposite, does not accommodate ‘no opinion’. Yet, he fails to recognize what went so wrong that there is no place in his spectrum for ‘no opinion’. The reason, that there is no place, is that he has identified a negative opinion as zero, rather than as -1. If he had identified a negative opinion as -1, which is the opposite of his +1 for a positive opinion, then ‘no opinion’ would have had a place in his spectrum of certitude at its midpoint of zero. Instead Dawkins discusses the distinction between temporarily *true* zeros in practice and permanently* true* zeros in principle, neither of which are accommodated by his spectrum ‘between two extremes of opposite certainty’ in which the opposite extreme of positive is not negative, but a *false* zero.

2) Is probability in the sense of rating one’s personal certitude of the truth of a statement a synonym for mathematical probability, which is the fractional concentration of an element in a logical set? For example, is probability used univocally in these two concepts: (a) The probability of a hurricane’s assembling a Boeing 747 while sweeping through a scrapyard. (b) The probability of a multiple of three in the set of integers, one through six?

Correct Answer: No. The probability of a hurricane’s assembling a Boeing 747 does not identify, even implicitly, a set of elements, one of which is the assembling of a Boeing 747. Probability as a numerical rating of one’s personal certitude of the truth of such a proposition has nothing to do with mathematical probability. In contrast to the certitude of one’s opinion about the capacity of hurricanes to assemble 747’s, the probability of a multiple of 3 in the set of integers, one through six, namely 1/3, is entirely objective.

Student Dawkins: Probability as the rating of one’s personal certitude of the truth of a proposition, has the same spectrum of definition as mathematical probability, namely 0 to +1 (p 50, Ref. 1). The odds against assembling a fully functioning horse, beetle or ostrich by randomly shuffling its parts are up there in the 747 territory of the chance that a hurricane, sweeping through a scrapyard would have the luck to assemble a Boeing 747. (p 113, Ref. 1) There is only one meaning of probability, whether it is the probability of the existence of God, the probability of a hurricane’s assembling a Boeing 747, the probability of success of Darwinian evolution based on the random generation of mutations or the probability of seven among the mutations of the sum of the paired output of two generators of random numbers to the base, six.

3) In arithmetic, is there a distinction between the factors of a product and the parts of a sum? Is the probability of a series of probabilities, the product or the sum of the probabilities of the series?

Correct Answers: Yes; The product. The individual probabilities of the series are its factors.

Student Dawkins: Yes; The relationship of a series of probabilities is more easily assessed from the perspective of improbability. An improbability can be broken up into smaller pieces of improbability. Those, who insist that the probability of a series is the product of the probabilities of the series, don’t understand the power of accumulation. Only an obscurantist would point out that if a large piece of improbability can be broken up into smaller pieces of improbability as the parts of a sum, i.e. as parts of an accumulation, then it must be true that its complement, the corresponding small piece of probability, is concomitantly broken up into larger pieces of probability, where the larger pieces of probability are the parts, whose sum (accumulation) equals the small piece of probability. (p 121, Ref. 1).

4) Jack and Jill go to a carnival. They view one gambling stand where for one dollar, the gambler can punch out one dot of a 100 dot card, where each dot is a hidden number from 00 to 99. The 100 numbers are randomly distributed among the dots of each card. If the gambler punches out the dot containing 00, he wins a kewpie doll. Later they view another stand where for one dollar, the gambler gets one red and one blue card, each with 10 dots. The hidden numbers 0 to 9 are randomly distributed among the dots of each card. If in one punch of each card, the gambler punches out red 0 and blue 0, he wins a kewpie doll. This second stand has an interesting twist, lacking in the first stand. A gambler, of course, may buy as many sets of one red card and one blue card as he pleases at one dollar per set. However, he need not pair up the cards to win a kewpie doll until after he punches all of the cards and examines the results.

(a) If a gambler buys one card from stand one and one pair of cards from stand two, what are his respective probabilities of winning a kewpie doll?

Correct Answer: The probability is 1/100 for both.

Student Dawkins: The probability of winning in one try at the first stand is 1/100. At the second stand the probability of winning is smeared out into probabilities of 1/10 for each of the two tries.

(b) How many dollars’ worth of cards must a gambler buy from each stand to reach a level of probability of roughly 50% that he will win at least one kewpie doll?

Correct Answers: $69 worth or 69 cards from stand one yields a probability of 50.0%. $12 worth or 24 cards (12 sets) from stand two yields a probability of 51.5%. (A probability closer to 50% for the second stand is not conveniently defined.)

Student Dawkins: A maximum of $50 and 50 cards from stand one yields a probability of 50%. A maximum of $49 and 49 cards from stand one yields a probability of 49%.A maximum of $14 and 28 cards yields a probability of 49%. (A probability closer to 50% for the second stand is not conveniently defined.)

(c) In the case described in (b), is the probability of winning greater at stand two?

Correct Answer: No

Student Dawkins: Yes

(d) In the case described in (b), is winning more efficient or less efficient in terms of dollars and in terms of total cards at the second carnival stand?

Correct Answer: More efficient. The second stand is based on two sub-stages of Darwinian evolution compared to the first stand, which is based on one overall stage of Darwinian evolution. The gradualism of sub-stages is more efficient in the number of random mutations while having no effect on the probability of evolutionary success. Efficiency is seen in the lower input of $12 or 24 random mutations compared to $69 or 69 random mutations to produce the same output, namely the probability of success of roughly 50%.

Student Dawkins: Efficiency is irrelevant. It’s all about probability. The gradualism of stand two breaks up the improbability of stand one into smaller pieces of improbability. (p 121, Ref. 1)

This problem is an illustration of two mutation sites of ten mutations each. I analyzed these relationships in Ref. 2, using an illustration of three mutation sites of six mutations each. In that illustration, I introduced two other modifications. One modification was that the winning number was unknown to the gambler. The other was that the gambler could choose the specific numbers on which to bet, so his tries or mutations were non-random. With the latter deviation from the Darwinian algorithm, the probability of winning a kewpie doll required a maximum of 216 non-random tries for the first stand and a maximum of 18 non-random tries for the second stand. The gradualism of the second stand smears out the luck required by the first stand. The increased probability of winning a kewpie doll at the second stand is due to the fact that one need not get his luck in one big dollop, as one does at the first stand. He can get it in dribs and drabs. It takes, respectively at the two stands, maxima of 216 and 18 tries for a probability of 100% of winning a kewpie doll. Consequently and respectively, it would take maxima of 125 and 15 tries to achieve a probability of 57.9% of winning a kewpie doll. Whether one compares 216 tries for the first stand to 18 tries for the second stand or 125 tries to 15 tries, the probability of winning a kewpie doll is greater at the second stand because it takes fewer tries. (See also the Wikipedia explanation, which is in agreement with Student Dawkins, Ref. 3)

Another example of extreme improbability is the combination lock of a bank vault. A bank robber could get lucky and hit upon the combination by chance. In practice the lock is designed with enough improbability to make this tantamount to impossible. But imagine a poorly designed lock. When each dial approaches its correct setting the vault door opens another chink. The burglar would home in on the jackpot in no time. ‘In no time’ indicates greater probability than that of his opening the well-designed lock. Any distinction between probability of success and efficiency in time is irrelevant. Also, any distinction between the probability of success and efficiency in tries, whether the tries are random mutations or non-random mutations is irrelevant. (p 122. Ref. 1)

5) If packages of 4 marbles each are based on a density of 1 blue marble per 2 marbles, how many blue marbles will a package selected at random contain?

Correct Answer: 2 blue marbles

Student Dawkins: 2 blue marbles.

6) If packages of 4 marbles each are based on a probability of blue marbles of 1/2, how many blue marbles will a package selected at random contain?

Correct Answer: Any number from 0 to 4 blue marbles.

Student Dawkins: 2 blue marbles. This conclusion is so surprising, I’ll say it again: 2 blue marbles. My calculation would predict that with the odds of success at 1 to 1, each package of 4 marbles would contain 2 blue marbles. (p 138, Ref. 1)

References:

1. *The God Delusion*

2. http://www.youtube.com/watch?v=JW1rVGgFzWU minute 4:25

3. http://en.wikipedia.org/wiki/Growing_Up_in_the_Universe#Part_3:_Climbing_Mount_Improbable

Calculations:

Where the total number of generated mutations, x, is random, and n is the number of different mutations, the Probability of success, P, equals 1- ((n – 1)/n)^x.

For n = 100 and x = 69, P = 50.0%

For n = 10 and x = 12, P = 71.7%. For P^2 = 51.5%, the sum of x = 24

Where the total number of generated mutations, x, is non-random, and n is the number of different mutations, the Probability of success, P, equals x/n

For n = 100 and x = 50, P = 50%

For n = 100 and x = 49, P = 49%

For n = 10 and x = 7, P = 70.0%. For P^2 = 49%, the sum of x = 14

For n = 216 and x = 216, P = 100%

For n = 6 and x = 6, P = 100%. For P^3 = 100%, the sum of x = 18

For n = 216 and x = 125, P = 57.9%

For n = 6 and x = 5, P = 83.3%. For P ^3 = 57.9%, the sum of x = 15